How to show $ a^2 <10^{\sqrt{a}}$ for $a\geq 2 $ and $ a \in \mathbb{N}$?

Should I try considering a new function which is the difference and then differentiating it?

I could solve it using two case when $ a \in [10^{2m}, 10^{2m+1})$ or $[10^{2m-1}, 10^{2m})$. Is there any other elementary way?

Any help would be appreciated. Thanks in advance.


2 Answers 2


That inequality is equivalent to $\ln a < \sqrt{a} \frac{\ln 10}{2}$ because of the fact that $t \to \ln t$ is strictly increasing.

Define $f(a)=\ln a$ and $g(a)=\sqrt{a} \frac{\ln 10}{2}$ for $a \geq 2$.

We want to show that $f(a)<g(a)$ for all $a \geq 2$.

Noting that $f'(a)=\frac 1 a \leq \frac {1}{\sqrt{a}}\frac {\ln 10} {4}=g'(a)$ holds if and only if $\sqrt{a} \geq \frac{4}{\ln 10}$, define $\phi(a)=g(a)-f(a)$.

Due to the fact that $\phi'(a) \geq 0$ if and only if $a \geq (\frac{4}{\ln 10})^2$ as we have observed and that $\phi((\frac{4}{\ln 10})^2)=g((\frac{4}{\ln 10})^2)-f((\frac{4}{\ln 10})^2)=2(1+\ln (\frac {\ln 10}{4})) >0$ we have what we want.

  • $\begingroup$ @mathisfun Yes the base of $\log$ is $e$. You shoul derivate $\log a$ and $\sqrt{a} \frac{\log 10}{2}$ to get that $\frac 1 a \leq \frac {1}{\sqrt{a}}\frac {\log 10} {4}$ $\endgroup$ Jun 14, 2020 at 12:09
  • $\begingroup$ Check the last line it is not correct..in order that the above inequality holds we need $ \sqrt{a} > \frac{4}{\ln 10}$ which means $ a>{(\frac{4}{\ln 10})}^2$ . Your last line is incorrect. $\endgroup$ Jun 14, 2020 at 15:48
  • $\begingroup$ @mathisfun I fixed it. Thank you for the remark. $\endgroup$ Jun 14, 2020 at 16:08

Since, $t \to \ln t$ is strictly increasing, it is equivalent to proving $\ln a < \sqrt{a}\cdot \frac{\ln 10}{2}$.
Let us define $\phi (x) =\sqrt{x}\cdot \frac{\ln 10}{2} - \ln x$ for $x\geq 2$.
$\phi '(x)= \frac {1}{\sqrt{x}}\frac {\ln 10} {4} - \frac{1}{x}.$ So, $\phi '(x) >0 \iff \frac {1}{\sqrt{x}}\frac {\ln 10} {4} > \frac{1}{x} \iff \sqrt{x} > \frac{4}{\ln 10}$
Let, $d= (\frac{4}{\ln 10})^2$
Now, for $x > d,$ we have $\sqrt{x} > \frac{4}{\ln 10} \implies \phi '(x) >0 \implies \phi $ is strictly increasing for $x>d.$
Similarly, $\phi$ is strictly decreasing for $2 \leq x<d$ i.e. $\phi$ attains its minimum at $x=d.$
Therefore, for $x \geq 2,$ we have $\phi (x) \geq \phi (d) = 2 -2 \ln (\frac{4}{\ln 10}) >0.$
This implies, for $x \geq 2,$ $\phi (x) >0 \implies \sqrt{x} \cdot \frac{\ln 10}{2} > \ln x.$

  • $\begingroup$ I think it is correct. $\endgroup$ Jun 14, 2020 at 17:00
  • $\begingroup$ Dear downvoter, kindly let me know if there is anything wrong in this proof.. otherwise please don't downvote. $\endgroup$ Jun 14, 2020 at 20:32

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